The equation of a circle $C$ is $x^2+y^2-16x-6y+48 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2-16x) + (y^2-6y) = -48$ $(x^2-16x+64) + (y^2-6y+9) = -48 + 64 + 9$ $(x-8)^{2} + (y-3)^{2} = 25 = 5^2$ Thus, $(h, k) = (8, 3)$ and $r = 5$.